Write the equation of the line that is perpendicular to the line 2x − 3y = 3 and passes through the point (−8, .a) y = two thirdsx − 10. b) y = two thirdsx + 10. c) y = −three halvesx − 10. d) y = −three halvesx + 10

First we need to find the slope of the original equation
2x - 3y = 3
-3y = -2x + 3
y = 2/3x - 1the slope here is 2/3.
However, we need a perpendicular line...so we need to use the negative reciprocal slope. All that means is flip the slope and change the sign.
slope 2/3flip3/2change the sign...-3/2.
so our perpendicular line will need a slope of -3/2.

y = mx + b
slope(m) = -3/2
(-8,2)...x = -8 and y = 2
now we sub and solve for b, the y int
2 = -3/2(-8) + b
2 = 12 + b
2 - 12 = b
-10 = b

so our perpendicular line is : y = -3/2x - 10

2x-3y=3
 minus 2x from both sides
-3y = 3 - 2x
divide by -3 both sides
y = (2/3)x - 1
 
so the slope of the line is 2/3
 the slope of the perpendicular line will be -3/2
 
y = -(3/2)x + b substitute x = -8, y = 2 to find b

hope this helps